Tensor Product of dual linear maps Suppose $V$ and $W$ are finite dimensional linear spaces and $V^*$ as well as $W^*$ are their appropriate linear duals.
Now let $f: V \to W$ and $g: V \to W$ be linear maps. Is the following identity correct?
$f^* \otimes g^* = (f \otimes g)^*$
That is the tensor product of the dual linear maps, is the linear dual of the tensor product of the maps.
Can't find this neither on the Wikipedia page of the tensor product, nor on the Wikipedia page of the dual linear maps. Therefore its properly wrong? Don't think so.
 A: The problem is that the two maps you are considering do not have the same domain and codomain:
$$f^*\otimes g^*:W^*\otimes W^*\to V^*\otimes V^*\quad\quad (f\otimes g)^*:(W\otimes W)^*\to(V\otimes V)^*$$
so they cannot possibly be equal as maps. However, we do have a canonical map 
$$\eta_W:W^*\otimes W^*\to (W\otimes W)^*,\quad (\eta_W(\phi\otimes\psi))(w_1\otimes w_2) = \phi(w_1)\psi(w_2)$$ (as explained in this question) which is an isomorphism in the finite dimensional case. Although $f^*\otimes g^*$ and $(f\otimes g)^*$ are not equal, I believe that the following is a factorization of $f^*\otimes g^*$:
$$W^*\otimes W^*\xrightarrow{\ \eta_W \ }(W\otimes W)^*\xrightarrow{\ (f\otimes g)^* \ }(V\otimes V)^*\xrightarrow{\ \eta_V^{-1} \ }V^*\otimes V^*.$$
To prove this, it suffices to show that the square:
$$\require{AMScd} \begin{CD}
W^*\otimes W^* @>{\eta_W}>> (W\otimes W)^*\\ @V{f^*\otimes g^*}VV @VV{(f\otimes g)^*}V\\
V^*\otimes V^* @>>{\eta_V}> (V\otimes V)^*
\end{CD}
$$
commutes. Let $\phi\otimes\psi\in W^*\otimes W^*$. We want to show that
$$((f\otimes g)^*\circ\eta_W)(\phi\otimes\psi) \quad\text{and}\quad ((f^*\otimes g^*)\circ\eta_V)(\phi\otimes\psi)$$
are equal in $(V\otimes V)^*$ (so equal as maps from $V\otimes V$ to the underlying field $\mathbb{F}$). With that in mind, let $v_1\otimes v_2\in V\otimes V$. Then, we have
\begin{align}
\notag ((f\otimes g)^*\circ\eta_W(\phi\otimes\psi))(v_1\otimes v_2) &= (\eta_W(\phi\otimes\psi)\circ(f\otimes g))(v_1\otimes v_2)\\
\notag &= \eta_W(\phi\otimes\psi)(f(v_1)\otimes g(v_2))\\
\notag &= \phi(f(v_1))\cdot \psi(g(v_2))
\end{align}
and:
\begin{align}
\notag (\eta_V\circ(f^*\otimes g^*)(\phi\otimes\psi))(v_1\otimes v_2) &= \eta_V((\phi\circ f)\otimes(\psi\circ g))(v_1\otimes v_2)\\
\notag &= \phi(f(v_1))\cdot \psi(g(v_2)).
\end{align}
Since everything was arbitrary, we have $f^*\otimes g^* = \eta_V^{-1}\circ (f\otimes g)^*\circ\eta_W$.
